system modeling in the time domain -...

ECE4510/5510: Feedback Control Systems. 2–1

SYSTEM MODELING IN THE TIME DOMAIN

2.1: What is a model? Why do we need one?

■ We use the term model to refer to a set of mathematical equations

used to represent a physical system, relating the system’s output

signal to its input signal.

■ A model is required in order to:

1. Understand system behavior (analysis).

2. Design a controller (synthesis).

KEY POINT: It is necessary to understand how the system works

naturally in order to know how to be able to change how it works

using a feedback controller.

■ Developing a reasonable mathematical model is the most important

part of the entire analysis.

■ It is also often the most difficult, amounting to ≈ 80 %–90 % of the

effort in designing a controller.

■ There are two basic approaches to modeling:

1. Analytic system modeling—we focus on these methods.

2. Empirical system identification. (In practice, there is always an

empirical component to system modeling: cf. ECE5560.)

■ It is important to realize that no model is ever exact! Inaccuracies

arise because of

Lecture notes prepared by and copyright c⃝ 1998–2017, Gregory L. Plett and M. Scott Trimboli

ECE4510/ECE5510, SYSTEM MODELING IN THE TIME DOMAIN 2–2

1. Unknown parameter values, or

2. Unmodeled dynamics (to make simpler model).

■ There is always a tradeoff between simplicity and accuracy:

• It’s often possible to improve the accuracy of a mathematical

model by increasing its complexity

• Simplification often means ignoring some inherent physical

properties

◆ e.g., ignore nonlinearities in linear, lumped-parameter models

• In general, it’s desirable to start with a simplified model to get a

“general feel,” increasing complexity only if the controlled system

does not meet performance requirements.

◆ Simplifications often ignore some high-frequency behaviors,

which requires that the controllers must operate with slower

transient-response requirements in order to be robust.

KEY POINT: “All models are wrong, but some are useful”

(George E. P. Box, statistician.)

EXAMPLE: Consider a 1 !, 2 W resistor.

■ Ohm’s Law (model) says: v(t) = i(t) · R.

i(t)

v(t)R

■ Apply 1 V. What happens?

• 1 A of current is predicted to flow.

• Power dissipated = V 2/R = 1 W.

■ Model should be accurate.



■ Now apply 10 V.

• 10 A of current is predicted to flow.

• Power dissipated = V 2/R = 100 W!

➤ Model will no longer be accurate.

➤ True behavior depends on input signal level—nonlinear.

➤ Model is accurate only in certain range of input-signal values.

■ Ohm’s law is definitely useful, but it is “wrong” in the sense that it

applies only under certain conditions, and even then is an averaged

version of what is truly happening at the microscopic scale.

LTI systems

■ This example shows that it is important to know the properties of your

model, as well as the model itself.

■ I claimed that the resistor exhibited “nonlinear” behavior, in some

sense.

■ In the next sections, we look at two critical properties of systems:

• Whether the system is “linear” (or not),

• Whether the system is “time invariant” (or not).

■ This course teaches methods to control linear-time-invariant (LTI)

systems.

■ Again, none exist! But, many are “close enough” for the techniques

developed here to work very well.

■ Also, we’ll look at ways of linearizing equations in Topic 2.7.



2.2: System properties of linearity and time invariance

TIME INVARIANT: The first system property that we look at is that of time

invariance.

■ A system is either time-varying or time-invariant, not both.

■ A time-invariant system does not change its fundamental behavior

over different periods of time. Its parameter values are constant.

■ A time invariant system satisfies the property (for any x(t), τ )

x(t − τ ) $→ y(t − τ )

when x(t) $→ y(t).

■ We can test a system for this property using ideas from the figure.

x1(t)

x2(t) = x1(t − τ )y2(t)

y1(t)y1(t − τ )

delay

delaysystem

system

■ A time-invariant system will have y2(t) = y1(t − τ ) for all x1(t) and τ .

TEST: To test for time-invariance, we must

■ Input x1(t) to the system and measure the output y1(t).

■ Input x2(t) = x1(t − τ ) to the system and measure y2(t).

■ If y2(t) = y1(t − τ ) for all possible delays τ and signals x1(t), then

the system is time-invariant.



EXAMPLE: For example, consider a square-law system y(t) = (x(t))2.

■ Input x1(t) to the system and measure y1(t): y1(t) = (x1(t))2 .

■ Input x2(t) to the system and measure y2(t): y2(t) = (x2(t))2 .

■ But, x2(t) = x1(t − τ ), so y2(t) = (x1(t − τ ))2 = y1(t − τ ).

■ Since this relationship holds for all τ and all x1(t), the square-law

system is time-invariant.

EXAMPLE: Let us examine a “delay operator.” (The delay operator is a

fundamental building-block of digital-signal-processing systems and

digital control systems).

■ The output of a delay is equal to the input, but shifted a constant

amount λ seconds: y(t) = x(t − λ), λ ≥ 0.

■ Input x1(t) to the system and measure y1(t): y1(t) = x1(t − λ).

■ Input x2(t) to the system and measure y2(t): y2(t) = x2(t − λ).

■ But, x2(t) = x1(t − τ ), so y2(t) = x1(t − τ − λ) = y1(t − τ ).

■ Since this relationship holds for all τ and all x1(t), the delay operator

is time-invariant.

EXAMPLE: Let us examine a “time compressor” whose output is equal to

its input, but “squashed” in time:y(t) = x(kt).

■ Input x1(t) to the system and measure y1(t): y1(t) = x1(kt).

■ Input x2(t) to the system and measure y2(t): y2(t) = x2(kt).

• Note that x2(kt) means that we take the input argument to the x2(·)

function and evaluate it at kt . That is, x2(kt) = x2(t)|t←kt .

■ But, x2(t) = x1(t − τ ). So, x2(kt) = x1(kt − τ ).



■ We want to know if y2(t) = x2(kt) = x1(kt − τ ) is equal to y1(t − τ ).

• Note that y1(t − τ ) means that we take the input argument to y1(·)

and evaluate it at t − τ . That is, y1(t − τ ) = x1(kt)|t←t−τ or

y1(t − τ ) = x1(k(t − τ )) = x1(kt − kτ ).

■ So, we have that y2(t) = x1(kt − τ ) = x1(kt − kτ ) = y1(t − τ ).

■ Therefore, the compressor is time-varying.

t

x1(t)

)⇒ t

y1(t)

t

x2(t) = x1(t − τ )

)⇒ t

y2(t)

= t

y1(t − τ )

LINEAR: The second property that we look at is linearity.

■ For linear systems, if x1(t) $→ y1(t) and x2(t) $→ y2(t), then

x3(t) = αx1(t) + βx2(t) $→ y3(t) = αy1(t) + βy2(t),

for any such x1(t), x2(t), α, β.



■ We can test a system for this property using ideas from the figure.

x1(t) x1(t)

x2(t) x2(t)

y1(t)

y2(t)

α α

ββ

x3(t)y3(t)y(t)

system

system

system

TEST: To test for linearity, we must

■ Input x1(t) to the system and measure the output y1(t).

■ Input x2(t) to the system and measure y2(t).

■ Input x3(t) = αx1(t) + βx2(t) to the system and measure y3(t).

■ If y3(t) = αy1(t) + βy2(t) for all possible α and β values, and x1(t)

and x2(t), then the system is linear.

EXAMPLE: Is the following system, described by the differential equation

y(t) + t y(t) = x(t), linear?1

■ Input x1(t) and output is y1(t): y1(t) + t y1(t) = x1(t).

■ Input x2(t) and output is y2(t): y2(t) + t y2(t) = x2(t).

■ Input x3(t) = αx1(t) + βx2(t) and measure y3(t).

y3(t) + t y3(t) = x3(t);

but, x3(t) = αx1(t) + βx2(t), so

y3(t) + t y3(t) = αx1(t) + βx2(t)

= α (y1(t) + t y1(t)) + β (y2(t) + t y2(t))

1 Note, the “dot” decoration on a variable indicates a time derivative. That is, y(t) =dy(t)/d(t), and so forth.



=d

dt(αy1(t) + βy2(t)) + t (αy1(t) + βy2(t)) .

By examining both sides of this equation, we realize that

y3(t) = αy1(t) + βy2(t). Therefore, the system is linear.

EXAMPLE: Trying this on the square-law system,

■ Input x1(t) and output is y1(t): y1(t) = (x1(t))2 .

■ Input x2(t) and output is y2(t): y2(t) = (x2(t))2 .

■ Input x3(t) = αx1(t) + βx2(t) and measure y3(t).

y3(t) = (x3(t))2

= (αx1(t) + βx2(t))2

= α2(x1(t))2 + 2αβx1(t)x2(t) + β2(x2(t))

2

= α (x1(t))2 + β (x2(t))

2 .

■ So, the square-law system is not linear (it is a nonlinear system).

KEY POINT: If a system is LTI, then it has an impulse response. This

entirely characterizes the system’s dynamics. The Laplace transform

of the impulse response is the transfer function. Working with the

transfer function eliminates the need to mess around with trying to

solve complicated differential equations.



2.3: Dynamics of mechanical systems (translational)

■ We now begin to review some basic physics as a refresher to

developing models of dynamic systems.

■ We’ll focus on mechanical and electrical systems, but will mention

some others too.

Translational motion

■ Newton’s second law, applied to translational motion, states:∑

F = ma.

■ That is, the vector sum of forces = mass of object times inertial

acceleration.

■ “Free-body diagrams” are a tool to apply this law.

EXAMPLE: Cruise control model.

■ Write the equations of motion for the speed and forward motion of a

car assuming that the engine imparts a forward force of u(t).

1. Assume rotational inertia of wheels is negligible.

2. Assume that friction is proportional to car’s speed (viscous friction).

∑

F = ma

u(t)− bx(t) = mx(t)

or, x(t) +b

mx(t) =

u(t)

m

bx(t) u(t)

x(t)

m

■ If the variable of interest is speed (v(t) = x(t)), not position,

v(t) +b

mv(t) =

u(t)

m



■ Notice that the differential equation has “output variables” on the left

of “=”, and “input variables” on the right.

IMPORTANT POINT: All of our models of dynamical systems will be

differential equations involving the input (e.g., u(t)) and its derivatives

and the output (e.g., y(t)) and its derivatives. No other signals

(intermediate variables) are allowed in our solutions.

EXAMPLE: Car suspension.

Each wheel in a car suspension system has a tire, shock absorber

and spring. Write the one-dimensional (vertical) equations of

motion for the car body and wheel.

■ “Quarter-car model”

Road Surface

Inertial Reference

kw

bks

m1

m2

x(t)

y(t)

r (t)

■ Free-body diagram:

ks(y(t)− x(t)) b( y(t)− x(t))

kw(x(t)− r (t)) ks(y(t)− x(t)) b( y(t)− x(t))

m1 m2x(t) y(t)

■ The force from the spring is proportional to its stretch. The force from

the shock absorber is proportional to the rate-of-change of its stretch.


ECE4510/ECE5510, SYSTEM MODELING IN THE TIME DOMAIN 2–11∑

F = ma

b (y(t)− x(t)) + ks (y(t)− x(t))− kw (x(t)− r(t)) = m1x(t)

−ks (y(t)− x(t))− b (y(t)− x(t)) = m2 y(t)

■ Re-arrange:

x(t) +b

m1

(x(t)− y(t)) +ks

m1

(x(t)− y(t)) +kw

m1

x(t) =kw

m1

r(t)

y(t) +b

m2

(y(t)− x(t)) +ks

m2

(y(t)− x(t)) = 0

Implementation in Simulink

■ Simulink is a component of MATLAB that is very useful for simulating

dynamic systems using a block-diagram approach.

■ Consider the cruise-control model, where we wish to control vehicle

velocity:

v(t) +b

mv(t) =

u(t)

m.

■ To implement this model in Simulink, we re-write the equation to have

only the highest derivative term on the left-hand-side

v(t) =u(t)

m−

b

mv(t).

■ We wire up a diagram like:



■ If the appropriate parameters are entered in the MATLAB workspace

for m and b, then this will simulate the car’s dynamics.

Important components for mechanical-translational systems:

x1(t)

x1(t)

x2(t)

x2(t)

f (t) = k(x1(t)− x2(t))

f (t) = b(x1(t)− x2(t))

k

b

m1. Mass

2. Spring

3. Damper



2.4: Dynamics of mechanical systems (rotational)

■ Newton’s second law, applied to rotational motion, states:∑

M = Jα or Iα.

■ That is, the vector sum of moments = moment of inertia times angular

acceleration. (“moment”=“torque”).

EXAMPLE: Satellites require attitude control so that sensors, antennas,

etc., are properly pointed. Let’s consider one axis of rotation.

θ(t)

d

Gas jetFc(t)

moment = Fc(t) · d, so,

Fc(t)d = J θ(t)

θ(t) =Fc(t)d

J

Note: Output of system θ(t) integrates

torques twice—“double-integrator plant.”EXAMPLE: A torsional pendulum is used, for example, in clocks enclosed

in glass domes. A similar device is the read-write head on a hard-disk

drive.

k

bJ

τ, θ

k: “Springiness” of suspension wire.

b: Viscous friction.

∑

M = J θ(t)

J θ(t) = τ (t)− bθ(t)− kθ(t)

θ(t) +b

Jθ(t) +

k

Jθ(t) =

τ (t)

J



Implementation in Simulink

■ The same basic principle applies to implementing this system in

Simulink as well, except now we have a second-derivative term.

■ No problem! Again, we re-write the equation to have only the highest

derivative term on the left-hand-side

θ(t) =τ (t)

J−

b

Jθ(t)−

k

Jθ(t).

■ We wire up a diagram like:

■ If the appropriate parameters are entered in the MATLAB workspace

for J , b, and k, then this will simulate the pendulum’s dynamics.



Important components for mechanical rotational systems:

θ1(t)

θ1(t)

θ2(t)

θ2(t)

τ (t) = k(θ1(t)− θ2(t))

τ (t) = b(θ1(t)− θ2(t))

k

b

J1. Inertia

2. Spring

3. Damper

Summary of Developing Models for Rigid Bodies:

1. Assign variables such as x(t) and θ(t) that are both necessary and

sufficient to describe and arbitrary position of the object.

2. Draw a free-body diagram of each component, and indicate all forces

acting on each body and the accelerations of the center of mass with

respect to an inertial reference.

3. Apply Newton’s laws:∑

F = ma,∑

M = Jα.

4. Combine the equations to eliminate internal forces.

5. The final form must be in terms of ONLY the input to the system and

its derivatives, and the output of the system and its derivatives.



2.5: Dynamics of electrical circuits

■ Kirchhoff’s Law’s:

• Current Law (KCL): The algebraic sum of currents entering a node

equals the algebraic sum of currents leaving the node.

• Voltage Law (KVL): The algebraic sum of all voltages taken around

a closed path in a circuit is zero.

■ “Node analysis” is a tool to apply these laws. (i.e., select one node as

reference (e.g., ground) and assume all other voltages are unknown.

Write equations for the unknowns using KCL. KVL must be used for

voltage sources.)

EXAMPLE: Bridged-T circuit.

R1 R2

C1

C2

vi(t)

➀ ➁ ➂

➃

■ Select reference = ➃.

• KVL at ➀: v➀(t) = vi(t).

• KCL at ➁:v➀(t)− v➁(t)

R1

−v➁(t)− v➂(t)

R2

− C1v➁(t) = 0.

• KCL at ➂:v➁(t)− v➂(t)

R2

+ C2(v➀(t)− v➂(t)) = 0.

v➁(t)− v➂(t) + R2C2(v➀(t)− v➂(t)) = 0

v➂(t) + R2C2(v➂(t)− v➀(t)) = v➁(t)



v➀(t)− [v➂(t) + R2C2(v➂(t)− v➀(t))]

R1

−

[v➂(t) + R2C2(v➂(t)− v➀(t))]− v➂(t)

R2

−

C1 [v➂(t) + R2C2(v➂(t)− v➀(t))] = 0

R2 (v➀(t)− [v➂(t) + R2C2(v➂(t)− v➀(t))])−

R1 ([v➂(t) + R2C2(v➂(t)− v➀(t))]− v➂(t))−

R1 R2C1 [v➂(t) + R2C2(v➂(t)− v➀(t))] = 0.

(

R1 R22C1C2

)

v➂(t) +(

R22C2 + R1 R2C2 + R1 R2C1

)

v➂(t) + (R2) v➂(t)

= (R1 R22C1C2)v➀(t) + (R2

2C2 + R1 R2C2)v➀(t) + (R2)v➀(t)

EXAMPLE: Op-amp circuit.

R1

R2 C

vi(t) vo(t)

i(t) =vi(t)

R1

vo(t) = −R2i(t)− vc(t).

dvo(t)

dt= −R2

di(t)

dt−

dvc(t)

dt

= −R2

R1

dvi(t)

dt−

i(t)

C

= −R2

R1

dvi(t)

dt−

1

R1Cvi(t)

R1C vo(t) = −R2C vi(t)− vi(t)

(as C →∞, we get an inverting amplifier.)



Important components for electrical systems:

1. Resistor

2. Capacitor

3. Inductor

4. Voltage source

5. Current source

6. Operational

Amplifier

v(t)

v(t)

v(t)

v(t)

i(t)

i(t)

i(t)

i(t)

vs

is

vd(t)i−(t)

i+(t) vo(t)

v(t) = Ri(t)

i(t) = Cd v(t)

dt

v(t) = Ld i(t)

dt

v(t) = vs

i(t) = is

vd(t) = 0

i−(t) = i+(t) = 0

vo(t) = Ao(v+(t)− v−(t))

as Ao→∞



2.6: Dynamics of electro-mechanical systems (etc.)

■ These are systems that convert energy from electrical to mechanical,

or vice versa.

EXAMPLE: DC generator.

■ Assume generator is driven at constant speed.

■ Generator has field windings (input), and rotor/armature windings

(output).R f Ra Lai f (t) ia(t)

e f (t) L f eg(t) Zlea(t)

︸︷︷︸

Fieldcircuit

︸︷︷︸

Armaturecircuit

︸︷︷︸

Loadcircuit

■ e f (t) = R f i f (t) + L f

di f (t)

dte f (t) is input, i f (t) is output.

■ eg(t) = Kφdθ(t)

dtK depends on generator structure.

= Kgi f (t). dθ(t)/dt = angular velocity = cst.

φ = flux, proportional to i f (t).

■ eg(t) = Raia(t) + La

dia(t)

dt+ Zlia(t). eg(t) is input, ia(t) is output.

■ ea(t) = Zlia(t). ia(t) is input, ea(t) is output.

e f (t) ea(t)i f (t) eg(t) ia(t)Field

circuitKg

Rotorcircuit

Zl

This is a preview of a block diagram used to simplify our understanding

of the system dynamics.



EXAMPLE: DC motor (servo-motor).

■ Directly generates rotational motion.

■ Indirectly generates translational motion.

ea(t) eb(t)

ia(t) Ra La b

Jθ(t), τ (t)

︸︷︷︸

Armature︸︷︷︸

Load

■ Mechanical resistance of load is translated into an electrical

“resistance” called the back e.m.f.

■ eb(t) = Ke

dθ(t)

dtKe = Kφ, as with generator.

■ ea(t) = Raia(t) + La

dia(t)

dt+ eb(t)

■ τ (t) = Kτ ia(t) Kτ = K1φ

■ Combining these equations of motion, recall Newton:∑

M = Jα

J θ(t) = τ (t)− bθ(t)

= Kτ ia(t)− bθ(t)

■ Assume (FOR NOW ONLY) electrical response is faster than

mechanical. La ≈ 0.

J θ(t) = Kτ

(

ea(t)− eb(t)

Ra

)

− bθ(t)

J θ(t) +

(

b +Kτ Ke

Ra

)

︸︷︷︸

back e.m.f. indistinguishable from friction!

θ(t) =Kτ

Ra

ea(t)



Dynamics of heat flow/ dynamics of fluid flow

■ These two subjects will not be covered here. Refer to texts on

thermodynamics or fluid-dynamics.

Transformers and gears

■ Ideally, both of these devices simply scale their input value.

Transformer :N1

N2

=e1

e2

=i2

i1

Gears :r1

r2

=θ2

θ1

=τ1

τ2

r1 r2

System identification (SYS ID)

■ When we generate models of system dynamics, we are performing

“system identifications.”

■ When we use known properties from physics and knowledge of the

system’s structure (as we have done here) we are performing “white

box system ID.”

■ If the system is very complex, or if the physics are not well

understood, we need to use input/output data to generate a system

model: “black-box system ID.”

■ A topic for the whole course! (ECE5560)



2.7: Linearization and analogous systems

■ We will study how to control linear systems.

■ Linear systems are rare.

■ We can “linearize” a non-linear system—the controller designed for

linearized model will work on the true nonlinear system (but not as

well as a controller designed directly for the non-linear system.)

EXAMPLE: NONLINEAR rotational pendulum.

l

θ(t), τ (t)

mg

Moment of inertia: J = ml2.∑

M = J θ(t)

J θ(t) = τ (t)− mgl sin(θ(t))

θ(t) +g

lsin(θ(t))︸︷︷︸

Nonlinear!

=τ (t)

ml2

■ If motion is “small,” sin (θ(t)) ≈ θ(t).

θ(t) +g

lθ(t) =

τ (t)

ml2Linear.

This is a preview of linearization.

KEY POINT: We can convert any differential equation into a first-order

vector differential equation:

˙x = f (x, u) ; x = vector, u = input.

Iff the system is linear, this will be of the form:

˙x = Ax + Bu; A and B are constant matrices.

EXAMPLE: Torsional pendulum revisited (pg. 2–13)

θ(t) +b

Jθ(t) +

k

Jθ(t) =

τ (t)

J



let

[

x1(t)

x2(t)

]

=

[

θ(t)

θ(t)

]

x2(t) +b

Jx2(t) +

k

Jx1(t) =

τ (t)

J[

x1(t)

x2(t)

]

=

[

0 1

−k/J −b/J

]

︸︷︷︸

A

[

x1(t)

x2(t)

]

+

[

0

1/J

]

︸︷︷︸

B

τ (t).

■ So, our model of the torsional pendulum is linear.

EXAMPLE: Rotational pendulum revisited (pg. 2–22)

θ(t) +g

lsin(θ(t)) =

τ (t)

ml2

let

[

x1(t)

x2(t)

]

=

[

θ(t)

θ(t)

]

[

x1(t)

x2(t)

]

=

⎡

⎣

x2(t)

−g

lsin(x1(t))

⎤

⎦ +

⎡

⎣

01

ml2

⎤

⎦ τ (t).

■ Not linear because we cannot make a constant A matrix.

Small-signal linearization

■ Uses a Taylor-series expansion of the differential equation around

some operating condition. (Equilibrium value where

x0 = 0 = f (x0, u0)).

let x = x0 + δx x0 = operating state

u = u0 + δu u0 = nominal control value.

x = f (x, u).



■ Taylor-series expansion:

x = x0 + δ x ≈ f (x0, u0) + Aδx + Bδu plus higher-order terms

■ Subtract out equilibrium (nominal) solution;

δ x = Aδx + Bδu,

which is linear. This is exactly how we linearized the rotational

pendulum before, with τ0 = 0; θ0 = 0.

sin(θ) = θ −θ3

3!+

θ5

5!− · · ·

≈ θ .

Feedback linearization (computed torque)

■ For rotational pendulum, ml2θ(t) + mgl sin(θ(t)) = τ (t).

• COMPUTE: τ (t) = mgl sin(θ(t)) + u(t).

• THEN: ml2θ(t) = u(t), no matter how large θ(t) becomes!

• Sometimes used in robotics and airplane flight control, but very

computationally intensive.

Analogous systems

■ The linearized differential equations of many very different physical

systems appear identical.

■ One would suppose they behave in similar ways (dynamic response)

and can be controlled with similar controllers.



Mechanical Translational mx(t) + bx(t) + kx(t) = u(t)

Mechanical Rotational J θ(t) + bθ(t) + kθ(t) = τ (t)

Satellite J θ(t) = f (t) · d

DC Motor (for La = 0) J θ(t) +

(

b +kτke

Ra

)

θ(t) =kτ

Ra

ea(t)

Generator (La L f )ea(t) + (L f (Ra + Rl) + La R f )ea(t)+

R f (Ra + Rl) = (kg Rl)e f (t)

■ These are all of the form

a2x(t) + a1x(t) + a0x(t) = b2u(t) + b1u(t) + b0u(t)

which is called a second-order form.

■ Therefore, we have seen very specific examples of a very general

class of system. If we learn how to control the general class, we can

apply this knowledge to specific systems.


system modeling in the time domain -...

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